Question: Find the number of 10-tuples $(x_1, x_2, \dots, x_{10})$ of real numbers such that
\[(1 - x_1)^2 + (x_1 - x_2)^2 + (x_2 - x_3)^2 + \dots + (x_9 - x_{10})^2 + x_{10}^2 = \frac{1}{11}.\]
Explanation: By the Cauchy-Schwarz inequality,
\begin{align*}
&[(1^2 + 1^2 + 1^2 + \dots + 1^2 + 1^2)][(1 - x_1)^2 + (x_1 - x_2)^2 + (x_2 - x_3)^2 + \dots + (x_9 - x_{10})^2 + x_{10}^2] \\
&\ge [(1 - x_1) + (x_1 - x_2) + (x_2 - x_3) + \dots + (x_9 - x_{10}) + x_{10}]^2 = 1.
\end{align*}From the given condition, we have equality, so by the equality condition for Cauchy-Schwarz,
\[\frac{1 - x_1}{1} = \frac{x_1 - x_2}{1} = \frac{x_2 - x_3}{1} = \dots = \frac{x_9 - x_{10}}{1} = \frac{x_{10}}{1}.\]Let
\[d = 1 - x_1 = x_1 - x_2 = x_2 - x_3 = \dots = x_9 - x_{10} = x_{10}.\]Then
\[(1 - x_1) + (x_1 - x_2) + \dots + (x_9 - x_{10}) + x_{10} = 11d,\]so $11d = 1.$  Then $d = \frac{1}{11},$ so
\[(x_1, x_2, x_3, \dots, x_{10}) = \left( \frac{10}{11}, \frac{9}{11}, \frac{8}{11}, \dots, \frac{1}{11} \right).\]In particular, there is only $\boxed{1}$ solution.